Matrix spaces

Row space

Range or column space

Null Space

Four subspaces Proof: https://www.math.tamu.edu/~dallen/linear_algebra/chpt4.pdf

Dimension Theorems

The dimension Theorem: Let V and U be vector spaces over the same field of scalars (either real numbers or complex numbers). Suppose that V is finite dimensional and let T be a linear transformation from V into U. Then

\[ \mbox{dim}\,\mbox{ker}(T) + \mbox{dim}\,\mbox{range}(T) = \mbox{dim}\,V . \]
Let n = dim V and k = dim ker(T), so 0 ≤ k ≤ n. If n = 0 or if k = n, there is nothing to prove, so we may assume that 0 ≤ k < n.

If k = 0, let α = {w1, w2, ... , wn} be a basis for V. Then

\[ \mbox{range}\left( T \right) = \mbox{span} \left\{ T{\bf w}_1 , T{\bf w}_2 , \ldots T{\bf w}_n \right\} , \]
so it suffices to show that the set β = {Tw1, Tw2, ... , Twn} is linearly independent. If
\[ {\bf 0} = c_1 T{\bf w}_1 + c_2 T{\bf w}_2 + \cdots + c_n T{\bf w}_n = T \left( c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_n {\bf w}_n \right) , \]
then c1w1 + c2w2 + ... + cnwn ∈ ker(T) = {0}, hence, the linear independence of α implies that c1 = c2 = ... = cn = 0 and so β is linearly independent.

If k ≥ 1, let {v1, v2, ... , vk} be a basis for ker(T), which we extend to a basis

\[ \alpha = \left\{ {\bf v}_1 , {\bf v}_2 , \ldots , {\bf v}_k , {\bf w}_1 , {\bf w}_2 , \ldots , {\bf w}_{n-k} \right\} \]
for V. Since Tv1 = Tv2 = ... = Tvk = 0, we have
\[ \mbox{range}(T) = \mbox{span} \left\{ T{\bf v}_1 , \ldots , T{\bf v}_k , T{\bf w}_1 , \ldots , T{\bf w}_{n-k} \right\} = \mbox{span} \left\{ T{\bf w}_1 , \ldots , T{\bf w}_{n-k} \right\} . \]
It suffices to show that β = {Tw1, Tw2, ... ,Twn-k} is linearly independent. If
\[ c_1 T{\bf w}_1 + c_2 T{\bf w}_2 + \cdots + c_{n-k} T{\bf w}_{n-k} = {\bf 0} , \]
then
\[ T \left( c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} \right) = {\bf 0} . \]
Therefore, c1w1 + c2w2 + ... + cn-kwn-k ∈ ker(T) and there are scalars a1, a2, ... , ak such that
\[ c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} = a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_k {\bf v}_k . \]
Then
\[ c_1 {\bf w}_1 + c_2 {\bf w}_2 + \cdots + c_{n-k} {\bf w}_{n-k} - a_1 {\bf v}_1 - a_2 {\bf v}_2 - \cdots - a_k {\bf v}_k = {\bf 0} , \]
so the linear independence of α implies that c1 = c2 = cn-k = a1 = a2 = ... = ak = 0. We conclude that β is linearly independent.

Corollary: Let V and U be finite dimensional vector spaces over the same field of scalars (either real numbers or complex numbers). Suppose that dimV = dim U and let T be a linear transformation from V into U. Then ker(T) = {0} if and only if the range of T is U.

If ker(T) = {0}, then dim ker(T) = 0 and dim range(T) = dimV = dim U. But the image of T is a subspace of U; hence range(T) = U.

Conversely, if range(T) = U, then dim range(T) = dimV = dim U and we get dim ker (T) = 0, so ker(T) = {0}.

Corollary (The dimension Theorem for matrices): Let A be an m×n matrix. Then

\[ \mbox{dim}\,\mbox{nullspace}({\bf A}) + \mbox{dim}\,\mbox{column}({\bf A}) = n \quad (\mbox{the number of columns in }\,{\bf A} . \]
If m = n, then the nullspace of A is {0} if and only if it is a full range matrix.
Apply the preceding theorem to the linear transformation generated by matrix A.